Thermodynamic Properties
Appendix A Thermodynamic Properties
Water
Molecular mass: M = 39.948 g/mol −3 Critical state parameters: pc = 221.2 bar, Tc = 647.3K,ρc = 17.54 × 10 mol/cm3 [1] Equilibrium liquid mass density [1]:
ρl = . + . 0.33 + . 3 mass 0 08 tanh y 0 7415 x 0 32 g/cm T x = 1 − , y = (T − 225)/46.2 Tc
Saturation vapor pressure [1]:
psat = exp [77.3491 − 7235.42465/T − 8.2lnT + 0.0057113 T ] Pa
Surface tension [1]:
−3 −3 2 γ∞ = 93.6635 + 9.133 × 10 T − 0.275 × 10 T mN/m
Lennard-Jones interaction parameters [2]:
σLJ = 2.641Å,εLJ/kB = 809.1K
Second virial coefficient [3]:
2 −3 3 B2(T ) = 17.1−102.9/(1−x) −33.6×10 (1−x) exp [5.255/(1 − x)] , cm /mol
V. I. Kalikmanov, Nucleation Theory, Lecture Notes in Physics 860, 293 DOI: 10.1007/978-90-481-3643-8, © Springer Science+Business Media Dordrecht 2013 294 Appendix A: Thermodynamic Properties
Nitrogen
Molecular mass: M = 28.0135 g/mol 3 Critical state parameters: pc = 33.958 bar, Tc = 126.192 K, ρc = 0.3133 g/cm [4] Equilibrium liquid mass density [4]:
l ρ . / ln sat = 1.48654237 x0 3294 − 0.280476066 x4 6 ρc / / + 0.0894143085 x16 6 − 0.119879866 x35 6, T x = 1 − Tc
Saturation vapor pressure [4]: psat Tc / / ln = −6.12445284 x + 1.2632722 x3 2 − 0.765910082 x5 2 − 1.77570564 x5 pc T
Surface tension [4]: 1.259 γ∞ = 29.324108 x mN/m
Lennard-Jones interaction parameters [2]:
σLJ = 3.798 Å,εLJ/kB = 71.4K
Pitzer’s acentric factor: ωP = 0.037 [2].
Mercury
Molecular mass: M = 200.61 g/mol −3 3 Critical state parameters: pc = 1510 bar, Tc = 1765 K, ρc = 23.41 × 10 mol/cm [5]. Saturation vapor pressure [5] a log p [Torr]=− + b + c lg T 10 sat T 10 with a = 3332.7, b = 10.5457, c =−0.848
Equilibrium liquid mass density [5]
ρl [ / 3]= . [ − −6( . + . 2 mass g cm 13 595 1 10 181 456 TCels 0 009205 TCels + . 3 + . 4 )] 0 000006608 TCels 0 000000067320 TCels Appendix A: Thermodynamic Properties 295 where TCels = T − 273.15 is the Celsius temperature. Surface tension [6] γ∞[mN/m]=479.4 − 0.22 TCels
Second virial coefficient [7]: TB ε1 ε2 B2 NA = c1 exp − 1 + c2 exp − − 1 T TB TB ε ε − c exp 1 − 1 + c exp − 2 − 1 cm3/mol (A.1) 1 T 2 T
3 where TB = 4286 K is the Boyle temperature of mercury, c1 = 69.87 cm /mol, 3 c2 = 22.425 cm /mol, ε1 = 655.8K,ε2 = 7563 K.
The value of coordination number N1 can be obtained from the measurements of the static structure factor. For fluid mercury it was studied over the whole liquid- vapor density range by Tamura and Hosokawa [8] and Hong et al. [9]usingX-ray diffraction measurements. Their data show that the first peak of the pair correlation function g(r) in the liquid phase is located at ≈ 3 Å and is relatively insensitive to the mass density in the range 10-13 g/cm3. The packing fraction in the liquid mercury at this range of densities is η ≈ 0.581. Using (7.68) one finds N1 ≈ 6.7. Argon
Molecular mass: M = 39.948 g/mol −3 Critical state parameters: pc = 48.6 bar, Tc = 150.633 K, ρc = 13.29 × 10 mol/cm3 [4] Equilibrium liquid mass density [4]: ρl = . + . 0.35 + . × 3 3, = − T mass M 13 290 24 49248 x 8 155083 x 10 g/cm x 1 Tc
Saturation vapor pressure [4]: p T . ln sat = c −5.904188529 x + 1.125495907 x1 5 − 0.7632579126 x3 − 1.697334376 x6 pc T
Surface tension [4]: 1.277 γ∞ = 37.78 x mN/m
Lennard-Jones interaction parameters [10]:
σLJ = 3.405 Å,εLJ/kB = 119.8K 296 Appendix A: Thermodynamic Properties
Pitzer’s acentric factor: ωP =−0.002 [2]. The second virial coefficient is given by the Tsanopoulos correlation for nonpolar substances Eq. (F.2).
N-nonane
Molecular mass: M = 128.259 g/mol −3 Critical state parameters: pc = 22.90 bar, Tc = 594.6K,ρc = 1.824 × 10 mol/cm3 [2] Equilibrium liquid mass density [2]:
ρl = . − . × −4 − . × −8 2 − . × −9 3 3 mass 0 733503 7 87562 10 TCels 9 68937 10 TCels 1 29616 10 TCels g/cm
where TCels = T − 273.15. Saturation vapor pressure [3]: −2 2 psat = exp −17.56832 ln T + 1.52556 10 T − 9467.4/T + 135.974 dyne/cm
Surface tension [3]:
γ∞ = 24.72 − 0.09347 TCels mN/m
The second virial coefficient [3]:
= . − . / + . / 2 − . / 3 − . / 8 3/ B2 NA 369 2 705 3 Tr 17 9 Tr 427 0 Tr 8 9 Tr cm mol where Tr = T/Tc. Appendix B Size of a Chain-Like Molecule
As one of the input parameters MKNT and CGNT use the size of the molecule. For a chain-like molecule, like nonane, it can be characterized by the radius of gyration Rg—the quantity used in polymer physics representing the mean square length between all pairs of segments in the chain [11]:
N segm 2 1 2 R = (Ri − R j ) g 2N 2 segm i, j=1 where Nsegm is the number of segments. Equivalently Rg can be rewritten as
N 1 segm R2 = (R − R )2 g N i 0 segm i=1 where R0 is the position of the center of mass of the chain. The latter expression shows that the chain-like molecule can be appropriately represented as a sphere with the radius Rg. The radius of gyration can be found using the Statistical Associating Fluid Theory (SAFT) [12]. Within the SAFT a molecule of a pure n-alkane can be modelled as a homonuclear chain with Nsegm segments of equal diameter σs and the same dispersive energy ε, bonded tangentially to form the chain. The soft-SAFT correlations for pure alkanes read [13]:
Nsegm = 0.0255 M + 0.628 (B.1) σ 3 = . + . Nsegm segm 1 73 M 22 8(B.2) where M is the molecular weight (in g/mol). Thus, the number of segments and the size of a single segment depend only on the molecular weight. Note, that within this approach Nsegm is not necessarily an integer number. Having determined Nsegm,the radius of gyration can be calculated using the Gaussian chain model in the theory of
V. I. Kalikmanov, Nucleation Theory, Lecture Notes in Physics 860, 297 DOI: 10.1007/978-90-481-3643-8, © Springer Science+Business Media Dordrecht 2013 298 Appendix B: Size of a Chain-Like Molecule polymers [11]: Nsegm Rg = σ (B.3) segm 6
Then, the effective diameter of the molecule can be estimated as
σ = 2 Rg (B.4)
For n-nonane Eqs. (B.1)–(B.2)give:
Rg = 3.202 Å,σ= 6.404 Å (B.5) Appendix C Spinodal Supersaturation for van der Waals Fluid
∗ v ∗ ∗ v In reduced units ρ = ρ /ρc, T = T/Tc, p = p /pc the van der Waals equation of state reads [14] ∗ ∗ ∗ ∗2 8ρ T p =−3ρ + (C.1) 3 − ρ∗
The spinodal equation ∂p∗/∂ρ∗ = 0is:
∗ ∗ ρ ∗ T = (3 − ρ )2 (C.2) 4 ρ∗v Solving Eq. (C.2) for the spinodal vapor density sp we obtain using the standard methods [15]: ∗v ∗ 1 ∗ ρ (T ) = 2 − 2 cos β ,β= arccos(1 − 2T ) (C.3) sp 3
Substitution of (C.3) into the van der Waals equation (C.1) yields the vapor pressure at the spinodal: ∗ 1 2 2 1 ∗v 8 4T − 3 cos β + 3 cos β sin β p = 3 3 6 (C.4) sp + 1 β 1 2 cos 3 from which the supersaturation at spinodal is
∗v ∗ p (T ) 1 S (T ∗) = sp = sp p∗ (T ∗) p∗ (T ∗) sat sat 8 4T ∗ − 3 cos 1 β + 3 cos 2 β sin2 1 β 3 3 6 (C.5) + 1 β 1 2 cos 3
V. I. Kalikmanov, Nucleation Theory, Lecture Notes in Physics 860, 299 DOI: 10.1007/978-90-481-3643-8, © Springer Science+Business Media Dordrecht 2013 Appendix D Partial Molecular Volumes
D.1 General Form
In this section we present a general framework for calculation of partial molecular volumes of components in a mixture. These quantities for a liquid phase are involved in the Kelvin equations. We consider here a general case of a two-phase m-component α, = ,..., ; α = , mixture vi i 1 m v l. The partial molecular volume of component i in the phase α is defined as ∂ α α = V , = , ,..., α = , vi α i 1 2 v l(D.1) ∂ N α, , α i p T N j, j=i where the quantities with the superscript α refer to the phase α. Since
N α V α = ρα
α the change of the total volume due to the change of Ni is α α 1 α N α dV = dN − dρ ρα i (ρα)2
α = α where we took into account that dN dNi . Then ⎡ ⎤ ∂ ρα α = 1 ⎣ − α ln ⎦ vi α 1 N α (D.2) ρ ∂ N α, , α i p T N j, j=i
The density ρα is an intensive property which can be expressed as a function of the set of intensive quantities—molar fractions of components in the phase α
V. I. Kalikmanov, Nucleation Theory, Lecture Notes in Physics 860, 301 DOI: 10.1007/978-90-481-3643-8, © Springer Science+Business Media Dordrecht 2013 302 Appendix D: Partial Molecular Volumes
N α α = j y j α (D.3) k Nk m α = ρα − α In view of normalization k=1 yk 1, is a function of m 1 variables yk and one is free to choose a particular component to be excluded from the list of independent variables. Discussing the partial molecular volume of component i, it is convenient to exclude this component from the list, i.e. to set
ρα = ρα( α,..., α , α ,..., ) y1 yi−1 yi+1 ym
Then the right-hand side of (D.2) can be expressed using the chain rule: α α ∂ α ∂ ln ρ ∂ ln ρ y j α = α α (D.4) ∂ N α, , α ∂y ∂ N i p T N j, j=i j=i j i
From (D.3) we find α α ∂y y j =− j ∂ α α (D.5) Ni v N N j, j=i
Substituting (D.5)into((D.2) and D.4) we obtain the general result α α 1 α α α ∂ ln ρ v = η ,η= 1 + y ,α= v, l(D.6) i ρα i i j ∂yα j=i j pα,T
In particular, for the binary a − b mixture (ya + yb = 1) we have: ∂ ρα ηα = − α ln a 1 yb α (D.7) ∂y α a p ,T ∂ ln ρα ηα = + yα b 1 a ∂ α (D.8) ya pα,T
Mention a useful identity resulting from (D.7)–(D.8):
ηα + ηα = ya a yb b 1(D.9)
D.2 Binary van der Waals Fluids
Here we present calculation of the partial molecular volumes of components in binary mixtures (vapor or liquid) described by the van der Waals equation of state: Appendix D: Partial Molecular Volumes 303 ρ kBT 2 p = − amρ (D.10) 1 − bmρ where the van der Waals parameters am, bm for the mixture read [2]: √ √ 2 am = (ya aa + yb ab) (D.11) bm = yaba + ybbb (D.12) where yi is the molar fraction of component i in vapor or liquid; ai and bi are the van der Waals parameters for the pure fluid i. According to the definition of the partial molecular volume consider a small perturbation in a number of molecules of component a at a fixed pressure, temperature and the number of molecules of component b. This perturbation results in the change of ρ, am and bm:
a m = am + Δam (D.13) bm = bm + Δbm (D.14) ρ = ρ + Δρ (D.15)
Substituting (D.13)–(D.15) into the van der Waals Eq. (D.10) and linearizing in Δam,Δbm and Δρ we find
ρk T Δρ k T ρ k T p − B + a ρ2 = B + B (Δb ρ + Δρ b ) − ρ2Δa − 2a ρΔρ m 2 m m m m 1 − bmρ 1 − bmρ (1 − bmρ)
The left-hand side vanishes in view of (D.10) resulting in Δb k T ρ2 (1 − b ρ)2 Δρ = A Δa − m B , A ≡ m 0 m 2 0 2 (D.16) (1 − bmρ) kBT − 2amρ(1 − bmρ)
Van der Waals parameters am and bm change due to the variation in molar fractions satisfying Δya =−Δyb: √ √ √ Δam = 2 Δya am ( aa − ab) (D.17) Δbm = Δya(ba − bb) (D.18)
Substituting (D.18)into(D.16) we obtain: √ √ √ (b − b ) k T Δρ = Δy A a ( a − a ) − a b B a 0 2 m a b 2 (D.19) (1 − bmρ)
Thus, for the binary van der Waals system 304 Appendix D: Partial Molecular Volumes
Table D.1 Reduced partial pv(bar) ηl ηl molecular volumes in the a b ηl , = , 1 1.005 0.289 liquid phase, i i a b,for the mixture n-nonane 10 1.057 0.305 (a)/methane (b) at T = 240 K 25 1.142 0.289 and various total pressures 40 1.228 0.365